department of eecs university of california, berkeley eecs 105 spring 2005, lecture 27 lecture 27:...

Department of EECS University of California, Berkeley

EECS 105 Spring 2005, Lecture 27

Lecture 27: PN Junctions

Prof. Niknejad


EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad

Diffusion

Diffusion occurs when there exists a concentration gradient

In the figure below, imagine that we fill the left chamber with a gas at temperate T

If we suddenly remove the divider, what happens? The gas will fill the entire volume of the new

chamber. How does this occur?



Diffusion (cont)

The net motion of gas molecules to the right chamber was due to the concentration gradient

If each particle moves on average left or right then eventually half will be in the right chamber

If the molecules were charged (or electrons), then there would be a net current flow

The diffusion current flows from high concentration to low concentration:



Diffusion Equations

Assume that the mean free path is λ Find flux of carriers crossing x=0 plane

)(n)0(n

)( n

0

thvn )(2

1 thvn )(2

1

)()(2

1 nnvF th

dx

dnn

dx

dnnvF th )0()0(

2

1

dx

dnvF th

dx

dnqvqFJ th



Einstein Relation

The thermal velocity is given by kT

kTvm thn 212*

21

cthv Mean Free Time

dx

dn

q

kTq

dx

dnqvJ nth

nn q

kTD

**2

n

c

n

ccthth m

q

q

kT

mkTvv



Total Current and Boundary Conditions

When both drift and diffusion are present, the total current is given by the sum:

In resistors, the carrier is approximately uniform and the second term is nearly zero

For currents flowing uniformly through an interface (no charge accumulation), the field is discontinous

dx

dnqDnEqJJJ nndiffdrift

21 JJ

2211 EE

1

2

2

1

E

E

)( 11 J

)( 22 J



Carrier Concentration and Potential

In thermal equilibrium, there are no external fields and we thus expect the electron and hole current densities to be zero:

dx

dnqDEqnJ o

nnn 000

dx

dn

kT

qEn

Ddx

dnoo

n

no 00

0

0

00 n

dnV

n

dn

q

kTd th

o



Carrier Concentration and Potential (2)

We have an equation relating the potential to the carrier concentration

If we integrate the above equation we have

We define the potential reference to be intrinsic Si:

)(

)(ln)()(

00

0000 xn

xnVxx th

inxnx )(0)( 0000

0

0

00 n

dnV

n

dn

q

kTd th

o



Carrier Concentration Versus Potential

The carrier concentration is thus a function of potential

Check that for zero potential, we have intrinsic carrier concentration (reference).

If we do a similar calculation for holes, we arrive at a similar equation

Note that the law of mass action is upheld

thVxienxn /)(

00)(

thVxienxp /)(

00)(

2/)(/)(200

00)()( iVxVx

i neenxpxn thth



The Doping Changes Potential

Due to the log nature of the potential, the potential changes linearly for exponential increase in doping:

Quick calculation aid: For a p-type concentration of 1016 cm-3, the potential is -360 mV

N-type materials have a positive potential with respect to intrinsic Si

100

0

0

0

00 10

)(log10lnmV26

)(

)(lnmV26

)(

)(ln)(

xn

xn

xn

xn

xnVx

iith

100

0 10

)(logmV60)(

xnx

100

0 10

)(logmV60)(

xpx



n-type

p-type

ND

NA

PN Junctions: Overview The most important device is a junction

between a p-type region and an n-type region When the junction is first formed, due to the

concentration gradient, mobile charges transfer near junction

Electrons leave n-type region and holes leave p-type region

These mobile carriers become minority carriers in new region (can’t penetrate far due to recombination)

Due to charge transfer, a voltage difference occurs between regions

This creates a field at the junction that causes drift currents to oppose the diffusion current

In thermal equilibrium, drift current and diffusion must balance

− − − − − −

+ + + + +

+ + + + ++ + + + +

− − − − − −− − − − − −

−V+



PN Junction Currents

Consider the PN junction in thermal equilibrium Again, the currents have to be zero, so we have

dx

dnqDEqnJ o

nnn 000

dx

dnqDEqn o

nn 00

dx

dn

nq

kT

ndxdn

DE

n

on

0

000

1

dx

dp

pq

kT

ndxdp

DE

p

op

0

000

1



PN Junction Fields

n-typep-type

NDNA

)(0 xpaNp 0

d

i

N

np

2

0 diffJ

0E

a

i

N

nn

2

0

Transition Region

diffJ

dNn 0

– – + +

0E

0px 0nx



Total Charge in Transition Region

To solve for the electric fields, we need to write down the charge density in the transition region:

In the p-side of the junction, there are very few electrons and only acceptors:

Since the hole concentration is decreasing on the p-side, the net charge is negative:

)()( 000 ad NNnpqx

)()( 00 aNpqx

0)(0 x0pNa

00 xxp



Charge on N-Side

Analogous to the p-side, the charge on the n-side is given by:

The net charge here is positive since:

)()( 00 dNnqx 00 nxx

0)(0 x0nNd

a

i

N

nn

2

0

Transition Region

diffJ

dNn 0

– – + +

0E



“Exact” Solution for Fields

Given the above approximations, we now have an expression for the charge density

We also have the following result from electrostatics

Notice that the potential appears on both sides of the equation… difficult problem to solve

A much simpler way to solve the problem…

0/)(

/)(

00)(

0)()(

0

0

nVx

id

poaVx

i

xxenNq

xxNenqx

th

th

s

x

dx

d

dx

dE

)(0

2

20



Depletion Approximation

Let’s assume that the transition region is completely depleted of free carriers (only immobile dopants exist)

Then the charge density is given by

The solution for electric field is now easy

0

0 0

0)(

nd

poa

xxqN

xxqNx

s

x

dx

dE

)(00

)(')'(

)( 000

00

p

x

xs

xEdxx

xEp

Field zero outsidetransition region



Depletion Approximation (2)

Since charge density is a constant

If we start from the n-side we get the following result

)(')'(

)(0

00 po

s

ax

xs

xxqN

dxx

xEp

)()()(')'(

)( 0000

00

0

xExxqN

xEdxx

xE ns

dx

xs

n

n

)()( 00 xxqN

xE ns

d

Field zero outsidetransition region



Plot of Fields In Depletion Region

E-Field zero outside of depletion region Note the asymmetrical depletion widths Which region has higher doping? Slope of E-Field larger in n-region. Why? Peak E-Field at junction. Why continuous?

n-typep-type

NDNA

– – – – –– – – – –– – – – –– – – – –

+ + + + +

+ + + + +

+ + + + +

+ + + + +

DepletionRegion

)()( 00 xxqN

xE ns

d

)()(0 pos

a xxqN

xE



Continuity of E-Field Across Junction

Recall that E-Field diverges on charge. For a sheet charge at the interface, the E-field could be discontinuous

In our case, the depletion region is only populated by a background density of fixed charges so the E-Field is continuous

What does this imply?

Total fixed charge in n-region equals fixed charge in p-region! Somewhat obvious result.

)0()0(00

xExqN

xqN

xE pno

s

dpo

s

an

nodpoa xqNxqN



Potential Across Junction

From our earlier calculation we know that the potential in the n-region is higher than p-region

The potential has to smoothly transition form high to low in crossing the junction

Physically, the potential difference is due to the charge transfer that occurs due to the concentration gradient

Let’s integrate the field to get the potential:

x

x pos

apo

p

dxxxqN

xx0

')'()()(

x

x

pos

ap

p

xxxqN

x

0

'2

')(

2



Potential Across Junction

We arrive at potential on p-side (parabolic)

Do integral on n-side

Potential must be continuous at interface (field finite at interface)

20 )(

2)( p

s

ap

po xx

qNx

20 )(

2)( n

s

dnn xx

qNx

)0(22

)0( 20

20 pp

s

apn

s

dnn x

qNx

qN



Solve for Depletion Lengths

We have two equations and two unknowns. We are finally in a position to solve for the depletion depths

20

20 22 p

s

apn

s

dn x

qNx

qN

nodpoa xqNxqN

(1)

(2)

da

a

d

bisno NN

N

qNx

2

ad

d

a

bispo NN

N

qNx

2

0 pnbi



Sanity Check

Does the above equation make sense? Let’s say we dope one side very highly. Then

physically we expect the depletion region width for the heavily doped side to approach zero:

Entire depletion width dropped across p-region

02

lim0

ad

d

d

bis

Nn NN

N

qNx

d

a

bis

ad

d

a

bis

Np qNNN

N

qNx

d

22lim0



Total Depletion Width

The sum of the depletion widths is the “space charge region”

This region is essentially depleted of all mobile charge

Due to high electric field, carriers move across region at velocity saturated speed

da

bisnpd NNq

xxX112

000

μ110

12150

qX bis

d

cm

V10

μ1

V1 4pnE



Have we invented a battery?

Can we harness the PN junction and turn it into a battery?

Numerical example:

2lnlnln

i

ADth

i

A

i

Dthpnbi n

NNV

n

N

n

NV

mV60010

1010logmV60lnmV26

20

1515

2

i

ADbi n

NN

?



Contact Potential

The contact between a PN junction creates a potential difference

Likewise, the contact between two dissimilar metals creates a potential difference (proportional to the difference between the work functions)

When a metal semiconductor junction is formed, a contact potential forms as well

If we short a PN junction, the sum of the voltages around the loop must be zero:

mnpmbi 0

pnmn

pm

+

−

bi )( mnpmbi



PN Junction Capacitor

Under thermal equilibrium, the PN junction does not draw any (much) current

But notice that a PN junction stores charge in the space charge region (transition region)

Since the device is storing charge, it’s acting like a capacitor

Positive charge is stored in the n-region, and negative charge is in the p-region:

nodpoa xqNxqN



Reverse Biased PN Junction

What happens if we “reverse-bias” the PN junction?

Since no current is flowing, the entire reverse biased potential is dropped across the transition region

To accommodate the extra potential, the charge in these regions must increase

If no current is flowing, the only way for the charge to increase is to grow (shrink) the depletion regions

+

−

Dbi V DV 0DV



Voltage Dependence of Depletion Width

Can redo the math but in the end we realize that the equations are the same except we replace the built-in potential with the effective reverse bias:

da

DbisDnDpDd NNq

VVxVxVX

11)(2)()()(

bi

Dn

da

a

d

DbisDn

Vx

NN

N

qN

VVx

1

)(2)( 0

bi

Dp

da

d

a

DbisDp

Vx

NN

N

qN

VVx

1

)(2)( 0

bi

DdDd

VXVX

1)( 0



Charge Versus Bias

As we increase the reverse bias, the depletion region grows to accommodate more charge

Charge is not a linear function of voltage This is a non-linear capacitor We can define a small signal capacitance for small

signals by breaking up the charge into two terms

bi

DaDpaDJ

VqNVxqNVQ

1)()(

)()()( DDJDDJ vqVQvVQ



Derivation of Small Signal Capacitance

From last lecture we found

Notice that

DV

DDJDDJ v

dV

dQVQvVQ

D

)()(

RD VVbipa

VV

jDjj

VxqN

dV

d

dV

dQVCC

1)( 0

bi

D

j

bi

Dbi

paj

V

C

V

xqNC

112

00

da

da

bi

s

da

d

a

bis

bi

a

bi

paj NN

NNq

NN

N

qN

qNxqNC

2

2

220

0



Physical Interpretation of Depletion Cap

Notice that the expression on the right-hand-side is just the depletion width in thermal equilibrium

This looks like a parallel plate capacitor!

da

da

bi

sj NN

NNqC

20

0

1

0

11

2 d

s

dabissj XNN

qC

)()(

Dd

sDj VX

VC



A Variable Capacitor (Varactor)

Capacitance varies versus bias:

Application: Radio Tuner

0j

j

C

C


EECS 105 Fall 2003, Lecture 27

Part II: Currents in PN Junctions



Diode under Thermal Equilibrium

Diffusion small since few carriers have enough energy to penetrate barrier Drift current is small since minority carriers are few and far between: Only

minority carriers generated within a diffusion length can contribute current Important Point: Minority drift current independent of barrier! Diffusion current strong (exponential) function of barrier

p-type n-type

DN AN

-

-

-

-

-

-

-

-

-

-

-

-

-

+++++++++++++

0E

biq

,p diffJ

,p driftJ

,n diffJ

,n driftJ

−

−

+

+

−

−

ThermalGeneration

Recombination Carrier with energybelow barrier height

Minority Carrier Close to Junction



Reverse Bias

Reverse Bias causes an increases barrier to diffusion

Diffusion current is reduced exponentially

Drift current does not change Net result: Small reverse current

p-type n-type

DN AN

-

-

-

-

-

-

-

+++++++

( )bi Rq V

+−



Forward Bias

Forward bias causes an exponential increase in the number of carriers with sufficient energy to penetrate barrier

Diffusion current increases exponentially

Drift current does not change Net result: Large forward current

p-type n-type

DN AN

-

-

-

-

-

-

-

+++++++

( )bi Rq V

+−



Diode I-V Curve

Diode IV relation is an exponential function

This exponential is due to the Boltzmann distribution of carriers versus

energy

For reverse bias the current saturations to the drift current due to minority

carriers

1dqV

kTd SI I e

dqV

kT

d

s

I

I

1

( )d d SI V I



Minority Carriers at Junction Edges

Minority carrier concentration at boundaries of depletion region increase as barrier lowers … the function is

)(

)(

pp

nn

xxp

xxp (minority) hole conc. on n-side of barrier

(majority) hole conc. on p-side of barrier

kTEnergyBarriere /)(

(Boltzmann’s Law)

kTVq DBe /)(

A

nn

N

xxp )(



“Law of the Junction”

Minority carrier concentrations at the edges of thedepletion region are given by:

kTVqAnn

DBeNxxp /)()(

kTVqDpp

DBeNxxn /)()(

Note 1: NA and ND are the majority carrier concentrations on the other side of the junction

Note 2: we can reduce these equations further by substituting

VD = 0 V (thermal equilibrium)Note 3: assumption that pn << ND and np << NA



Minority Carrier Concentration

The minority carrier concentration in the bulk region for forward bias is a decaying exponential due to recombination

p side n side

-Wp Wn xn -xp

0

AqV

kTnp e

0 0( ) 1A

p

xqVLkT

n n np x p p e e

0np0pn

0

AqV

kTpn e

Minority CarrierDiffusion Length



Steady-State Concentrations

Assume that none of the diffusing holes and electrons recombine get straight lines …

p side n side

-Wp Wn xn -xp

0

AqV

kTnp e

0np0pn

0

AqV

kTpn e

This also happens if the minority carrier diffusion lengths are much larger than Wn,p

, ,n p n pL W



Diode Current Densities

0 1A

p

qVpdiff n kT

n n ppx x

dn DJ qD q n e

dx W

0 0( )( )

AqV

kTp p p

p p

dn n e nx

dx x W

p side n side

-Wp Wn xn -xp

0

AqV

kTnp e

0np

0pn

0

AqV

kTpn e

0 1A

n

qVpdiff n kT

p p nx x n

DdpJ qD q p e

dx W

2 1AqV

pdiff n kTi

d n a p

D DJ qn e

N W N W

2

0i

pa

nn

N



Diode Small Signal Model

The I-V relation of a diode can be linearized

( )

1d d d dq V v qV qv

kT kT kTD D S SI i I e I e e

( )1 d d

D D D

q V vI i I

kT

2 3

12! 3!

x x xe x

dD d d

qvi g v

kT



Diode Capacitance

We have already seen that a reverse biased diode acts like a capacitor since the depletion region grows and shrinks in response to the applied field. the capacitance in forward bias is given by

But another charge storage mechanism comes into play in forward bias

Minority carriers injected into p and n regions “stay” in each region for a while

On average additional charge is stored in diode

01.4Sj j

dep

C A CX



Charge Storage

Increasing forward bias increases minority charge density By charge neutrality, the source voltage must supply equal

and opposite charge A detailed analysis yields:

p side n side

-Wp Wn xn -xp

( )

0

d dq V v

kTnp e

0np0pn

( )

0

d dq V v

kTpn e

1

2d

d

qIC

kT

Time to cross junction(or minority carrier lifetime)



Forward Bias Equivalent Circuit

Equivalent circuit has three non-linear elements: forward conductance, junction cap, and diffusion cap.

Diff cap and conductance proportional to DC current flowing through diode.

Junction cap proportional to junction voltage.



Fabrication of IC Diodes

Start with p-type substrate Create n-well to house diode p and n+ diffusion regions are the cathode and annode N-well must be reverse biased from substrate Parasitic resistance due to well resistance

p-type

p+

n-well

p-type

n+

annodecathode

p



Diode Circuits

Rectifier (AC to DC conversion) Average value circuit Peak detector (AM demodulator) DC restorer Voltage doubler / quadrupler /…

department of eecs university of california, berkeley eecs 105 spring 2005, lecture 27 lecture 27:...

Documents